Here is Prob. 3, Exercises 8.14, in the book Calculus Vol. II by Tom M. Apostol, 2nd edition:
Find the points $(x, y)$ and the directions for which the directional derivative of $f(x, y) = 3x^2+y^2$ has its largest value, if $(x, y)$ is restricted to be on the circle $x^2 + y^2 = 1$.
My Attempt:
We note that, at any point $(x, y) \in \mathbb{R}^2$, we have $$ \frac{ \partial f (x, y)}{ \partial x} = 6x, \qquad \mbox{ and } \qquad \frac{ \partial f (x, y)}{ \partial y } = 2y. $$ Thus these partial derivatives exist and are continuous everywhere in $\mathbb{R}^2$. Hence $f$ is differentiable everywhere in $\mathbb{R}^2$. Therefore we must have $$ f^\prime( (x, y); (u, v) ) = \nabla f(x, y) \cdot (u, v) = ( 6x, 2y) \cdot (u, v) = 6xu + 2yv $$ for any point $(x, y) \in \mathbb{R}^2$ and for every direction $(u, v) \in \mathbb{R}^2$.
Now at any point $(x, y) \in \mathbb{R}^2$, the derivative $f^\prime( (x, y) ; (u, v) )$ is largest in the direction of the gradient vector $(6x, 2y)$ and this largest value is $$ \lVert \nabla f(x, y) \rVert^2 = 36x^2 + 4y^2, $$ and, for all points $(x, y) \in \mathbb{R}^2$ for which $x^2 + y^2 = 1$, this largest value is $$ \lVert \nabla f(x, y) \rVert^2 = 36x^2 + 4y^2 = 36 \left(x^2 + y^2 \right) - 32 y^2 = 36 - 32 y^2, $$ and this last expression attains its maximum value when $y = 0$, that is, at either of the points $(1, 0)$ and $(-1, 0)$.
Thus the largest value of $f^\prime( (x, y); (u, v) )$ for any points $(x, y) \in \mathbb{R}^2$ for which $x^2+y^2=1$ is attained at the points $( 1, 0)$ and $(-1, 0)$ in the direction of the vectors $$\nabla f(1, 0) = ( 6, 0) $$ and $$\nabla f(-1, 0) = (-6, 0),$$ respectively, and this largest value in either case is $36$.
Is my solution correct? Is each and every step in it correct and clear enough? If not, then where are the issues?